Straight Enough Lines

Geometry Level 4

External angle bisectors of Angle B and Angle C are y = x and y = -2x respectively. If vertex A is (1 , 3) then find the co-ordinates of the incentre of the triangle .

(-1/2 , 1) (1 , 1) (1/2 , 3/2 ) (0, 0)

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3 solutions

Two external bisectors meet at I A ( 0 , 0 ) . I w i l l b e o n I A A . Y I A A = 3 X . O u t o f g i v e n p o i n t s o n l y ( 1 / 2 , 3 / 2 ) i s o n Y I A A = 3 X . I_A (0,0). \ \ I \ will\ be\ on \ I_AA.\ \ \therefore\ Y_{I_AA}=3X.\ Out\ of \ given\ points\ only \ (1/2,3/2) \ is\ on\ Y_{I_AA}=3X.

Ujjwal Rane
Oct 7, 2016

This can be solved by applying Fermat's principle (light takes the quickest path) to the path a ray of light would take to make a round trip starting from point A (the source) go to line y = x (a mirror) and line y = -2x (another mirror). This path is nothing but the desired triangle ABC!

To get it, reflect point A(1,3) about the two mirrors and join these images. Wherever the line intersects the two mirrors are the points B and C. The images A are (3,1) and (-3,1) Clearly, line y = 1 joins them, giving intercepts (1,1) and (-1/2,1) as the vertices of the desired triangle. From which the incenter can be worked out to be (1/2,3/2).

As normal with you, another out of box solution. Thank you. +1).

Niranjan Khanderia - 4 years, 8 months ago

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Thank you Sir!

Ujjwal Rane - 4 years, 8 months ago
Ayush Parasar
Jul 29, 2015

A simple solution of the problem : Please try and understand from it . If you have any further doubts or queries please ask .

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